LeetCode - Binary Tree Level Order Traversal

这题算是树遍历的基础，写这篇博客主要是为了记录两种代码风格的层序遍历。

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).For example:Given binary tree [3,9,20,null,null,15,7],    3   / \  9  20    /  \   15   7return its level order traversal as:[  [3],  [9,20],  [15,7]]

题目链接： https://leetcode.com/problems/binary-tree-level-order-traversal/

风格一：
记录当前层与下一层的节点数，一个队列保存节点，避免多队列的拷贝。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:       vector<vector<int>> levelOrderBottom(TreeNode* root) {        vector<vector<int>> res;        if(!root)            return res;        vector<int> vlevel;        queue<TreeNode*> q;        q.push(root);        int level1=1,level2=0;        while(!q.empty()){            TreeNode* front = q.front();            q.pop();            vlevel.push_back(front->val);            if(front->left){                q.push(front->left);                ++level2;            }            if(front->right){                q.push(front->right);                ++level2;            }            if(--level1 == 0){//one level is over                res.push_back(vlevel);                level1 = level2;//next level                level2=0;//next next level start                vlevel.clear();            }        }        reverse(res.begin(),res.end());        return res;    }};

风格二：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<vector<int>> levelOrder(TreeNode* root) {        vector<vector<int> > ret;        if(root == NULL)            return ret;        queue<TreeNode *> Q;        Q.push(root);        while(!Q.empty()) {            queue<TreeNode *> Qtmp;            vector<int> Vtmp;            while(!Q.empty()) {                TreeNode *front = Q.front();                Q.pop();                Vtmp.push_back(front->val);                if(front->left != NULL)                    Qtmp.push(front->left);                if(front->right != NULL)                    Qtmp.push(front->right);            }            ret.push_back(Vtmp);            Q = Qtmp;        }        return ret;    }};